Question

Question 1

In this unit, you learned about common logarithms. Since most calculators can be used to find the values of common logarithms only, it’s desirable to be able to convert logarithms with bases other than 10 to common logarithms. This conversion can be done using the change of base formula.


Part A

Rewrite the logarithmic equation y = logbm in exponential form.

Part B

Using a log with base c , take the log of both sides of your answer from part A. Then use the power property and solve for y .

Part C

The original equation was y=logbm . Use substitution and your answer from part B to write the change of base formula.

Part D

Watch this video for two examples of how to use the change of base formula to simplify and evaluate logarithmic expressions. Then simplify (or evaluate, if possible) this expression: log3z x logz27

Part E

Use the change of base formula and a calculator to evaluate log7 300. Round your ans

Answer 1

Answer and Step-by-step explanation:

Part A: Exponential equation is the "opposite" of logarithmic equation, so:

`y=log_(b)m`

`b^(y)=m`

Part B: Using log with base c:

`log_(c)b^(y)=log_(c)m`

Power property of logarithm states that if the anti-logarithm is elevated at a power, the elevated number can be pulled in front of the logarithm:

`ylog_(c)b=log_(c)m`

Solving for y:

`y=(log_(c)m)/(log_(c)b)`

Part C: To facilitate the use of calculators, which only have values for the base-10 log and natural log, we use change of base formula, i.e., transform

`y=log_(b)m`

into

`y=(log_(c)m)/(log_(c)b)`

Part D:
`(log_(3)z)(log_(z)27)`

Change of base will be:

`log_(3)z=(log_(10)z)/(log_(10)3)`

`log_(z)27=(log_(10)27)/(log_(10)z)`

Solving:

`(log_(3)z)(log_(z)27)` =
`((log_(10)z)/(log_(10)3))((log_(10)27)/(log_(10)z) )`

`(log_(3)z)(log_(z)27)` =
`(log_(10)27)/(log_(10)3)`

`(log_(3)z)(log_(z)27)` =
`(log_(10)3^(3))/(log_(10)3)`

`(log_(3)z)(log_(z)27)` =
`(3log_(10)3)/(log_(10)3)`

`(log_(3)z)(log_(z)27)` = 3

Part E:
`log_(7)300`

Using change of base:

`log_(7)300=(log300)/(log7)`

`log_(7)300=(2.48)/(0.85)`

`log_(7)300` ≈ 3